Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t - 2}{-2t^2 - 6t + 20} \div \dfrac{t^2 - 10t}{t^3 + 9t^2 + 20t} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{t - 2}{-2t^2 - 6t + 20} \times \dfrac{t^3 + 9t^2 + 20t}{t^2 - 10t} $ First factor out any common factors. $x = \dfrac{t - 2}{-2(t^2 + 3t - 10)} \times \dfrac{t(t^2 + 9t + 20)}{t(t - 10)} $ Then factor the quadratic expressions. $x = \dfrac {t - 2} {-2(t + 5)(t - 2)} \times \dfrac {t(t + 5)(t + 4)} {t(t - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(t - 2) \times t(t + 5)(t + 4) } { -2(t + 5)(t - 2) \times t(t - 10)} $ $x = \dfrac {t(t + 5)(t + 4)(t - 2)} {-2t(t + 5)(t - 2)(t - 10)} $ Notice that $(t + 5)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {t\cancel{(t + 5)}(t + 4)(t - 2)} {-2t\cancel{(t + 5)}(t - 2)(t - 10)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $x = \dfrac {t\cancel{(t + 5)}(t + 4)\cancel{(t - 2)}} {-2t\cancel{(t + 5)}\cancel{(t - 2)}(t - 10)} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $x = \dfrac {t(t + 4)} {-2t(t - 10)} $ $ x = \dfrac{-(t + 4)}{2(t - 10)}; t \neq -5; t \neq 2 $